# Thread: partial fractions

1. ## partial fractions

I am not sure how to express this as partial fractions. Never come across a cubic denominator before!

Q express $\frac{x^5-1}{x^2(x^3+1)}$ in partial fractions ?

2. Originally Posted by oxrigby
I am not sure how to express this as partial fractions. Never come across a cubic denominator before!

Q express $\frac{x^5-1}{x^2(x^3+1)}$ in partial fractions ?
$(x^3+1)=(x+1)(x^2-x+1)$

RonL

3. $\frac{x^5-1}{x^5+x^2}$

$\frac{x^5-1+x^2-x^2}{x^5+x^2}$

$\frac{x^5+x^2}{x^5+x^2}-\frac{1+x^2}{x^5+x^2}$

$1-\frac{1+x^2}{x^2(x^3+1)}$

I'd start it that way. You can try and finish it.

In this case:

$1+x^2\equiv(Ax+B)(x^3+1)+(Cx^2+Dx+E)(x^2)$

Since this is an identity, it is true for all values of x (this is a great help in determing the values of A,B,C,D and E!).

4. In general the grade of the polynomial in the numerator should be less than the grade of the polynomial in the denominator.

If it's not presented in that way, you should divide the numerator polynomial by the denominator and express it that way.

No tricks this way.

5. ## partial fractions

I was going back over this question and i wanted to express $\frac{1}{x^4-4x^2+16}$ in partial fractions having expressed $x^4-4x^2+16=(x^2+2\sqrt{3}+4)(x^2-2\sqrt{3}+4))$ quite worried actually as i could do it first time round by putting $\frac{1}{x^4-4x^2+16}=\frac{Ax+B}{x^2+2\sqrt{3}+4}+\frac{Cx+D}{ x^2-2\sqrt{3}+4}$ as you would expect. At this stage however I can only get 1=4B+4D. I previously got B=1/8=D and $A=\frac{1}{16\sqrt{3}}$ $C=\frac{-1}{16\sqrt{3}}$ as i said i cant get these this time round so if anyone could show me this would be great!

6. Originally Posted by oxrigby
I was going back over this question and i wanted to express $\frac{1}{x^4-4x^2+16}$ in partial fractions having expressed $x^4-4x^2+16=(x^2+2\sqrt{3}+4)(x^2-2\sqrt{3}+4))$ quite worried actually as i could do it first time round by putting $\frac{1}{x^4-4x^2+16}=\frac{Ax+B}{x^2+2\sqrt{3}+4}+\frac{Cx+D}{ x^2-2\sqrt{3}+4}$ as you would expect. At this stage however I can only get 1=4B+4D. I previously got B=1/8=D and $A=\frac{1}{16\sqrt{3}}$ $C=\frac{-1}{16\sqrt{3}}$ as i said i cant get these this time round so if anyone could show me this would be great!
I think you've made a mistake, $(x^2+2\sqrt{3}+4)(x^2-2\sqrt{3}+4))=x^4+8x^2+4$

$x^4-4x^2+16=(x+2e^{\frac{i\pi}{6}})(x-2e^{\frac{i\pi}{6}})(x+2e^{\frac{i5\pi}{6}})(x-2e^{\frac{i5\pi}{6}})$

I haven't worked through it further, I'd need to pull out my notes from last semester. If you have more trouble I'll see if I can remember what to do. Those are pi in the exponent by the way, they are so small on my screen they almost look like x.

7. i don't think ive got a mistake. I managed to get the answer which is $\frac{x+2\sqrt{3}}{16\sqrt{3}(x^2+2\sqrt{3}x+4}-\frac{x-2\sqrt{3}}{16\sqrt{3}(x^2-2\sqrt{3}+4)}$ a long time ago, the only thing i can derive now though is the 4B+4D=1 and i cant remeber how i go the 8 for A and B etc something has completely slipped my mind.

8. Originally Posted by oxrigby
i don't think ive got a mistake. I managed to get the answer which is $\frac{x+2\sqrt{3}}{16\sqrt{3}(x^2+2\sqrt{3}x+4}-\frac{x-2\sqrt{3}}{16\sqrt{3}(x^2-2\sqrt{3}+4)}$ a long time ago, the only thing i can derive now though is the 4B+4D=1 and i cant remeber how i go the 8 for A and B etc something has completely slipped my mind.
I think I see, you didn't have your x in the $2x\sqrt{3}$ terms when you factored originally so that confused me. You also left it out in the denominator of the second term of the previous post, although I'm not entirely sure where you got $x^4-4x^2+16$ to begin with.

9. Originally Posted by Showcase_22
$\frac{x^5-1}{x^5+x^2}$

$\frac{x^5-1+x^2-x^2}{x^5+x^2}$

$\frac{x^5+x^2}{x^5+x^2}-\frac{1+x^2}{x^5+x^2}$

$1-\frac{1+x^2}{x^2(x^3+1)}$

I'd start it that way. You can try and finish it.

In this case:

$1+x^2\equiv(Ax+B)(x^3+1)+(Cx^2+Dx+E)(x^2)$

Since this is an identity, it is true for all values of x (this is a great help in determing the values of A,B,C,D and E!).
But the whole point of the question is that any cubic can be factored further. In this case, $x^3+ 1= (x+ 1)(x^2- x+ 1)$ as Captain Black said in the very first response to this question.

10. lol apologies that $\frac{x^5-1}{x^2(x^3+1)}$ is an entirely different question which I understand the principles of now. Yes you are right i missed the x. the question was express $x^4-4x^2+16 in form (x^2+Ax+B)(x^2+Cx+D)$ which i did by completing the square(matching coefficients didnt seem to work!) so I get the answer then it says hence express $\frac{1}{x^4-4x^2+16}$ in partial fractions(this is where i get stuck) i do = $\frac{Ax+B}{x^2+2\sqrt{3}x+4}=\frac{Cx+D}{x^2-2\sqrt{3}x+4}$ the answer is what i said in my other post with root 3 on the denominator but i cant figure from this point on wards how i did this a long time ago,,,i can only get 4B+4D=1 putting in x=0 any suggestions??? regards